The energy of a particle executing simple har | vedclass

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Question

(A) The total energy of a particle in $SHM$ is $E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$. Comparing this with $E = Ax^2 + Bv^2$,we get $A = \frac{k}{2}$

Vedclass · Accepted Answer

Amplitude of $SHM$ is $\sqrt{\frac{E}{A}}$

Vedclass · Answer

(A) The total energy of a particle in $SHM$ is $E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$. Comparing this with $E = Ax^2 + Bv^2$,we get $A = \frac{k}{2}$ and $B = \frac{m}{2}$.$1$. At maximum displacement $(x = A_{amp})$,velocity $v = 0$. Thus,$E = A(A_{amp})^2$,which gives $A_{amp} = \sqrt{\frac{E}{A}}$. (Option $A$ is correct).$2$. At mean position $(x = 0)$,velocity is maximum $(v_{max})$. Thus,$E = B(v_{max})^2$,which gives $v_{max} = \sqrt{\frac{E}{B}}$. (Option $B$ is correct).$3$. Angular frequency $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{2A}{2B}} = \sqrt{\frac{A}{B}}$. The time period $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{B}{A}}$. (Option $C$ is correct).$4$. Maximum acceleration $a_{max} = \omega^2 A_{amp} = \left(\frac{A}{B}\right) \sqrt{\frac{E}{A}} = \sqrt{\frac{A^2 E}{B^2 A}} = \frac{\sqrt{AE}}{B}$.Wait,checking option $D$: $\frac{\sqrt{EA}}{B}$ is indeed the correct expression for maximum acceleration. Re-evaluating the question: The amplitude in option $A$ was given as $\sqrt{\frac{2E}{A}}$ in the prompt,which is incorrect. Therefore,option $A$ is the incorrect statement.

The energy of a particle executing simple harmonic motion is given by $E = Ax^2 + Bv^2$,where $x$ is the displacement from mean position $x = 0$ and $v$ is the velocity of the particle at $x$. Choose the incorrect statement.

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